Power to Karl!
We are told that the equation 1 8 1 8 = a 2 + b 2 has one and only one integer solution with 0 < a < b . Find a + b without the use of a calculator.
We dedicate this simple problem to the birthday of the great Karl Marx, who was born on May 5, 1818.
The answer is 60.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
Moderator note:
What a pleasant surprise! Superb observation.
Another delightfil solution; thanks! As you know, the identity you use is just the product rule for the modulus of complex numbers: ( a 2 + b 2 ) ( c 2 + d 2 ) = ∣ a + b i ∣ 2 ∣ c + d i ∣ 2 = ∣ a c − b d + ( a d + b c ) i ∣ 2 = ( a c − b d ) 2 + ( a d + b c ) 2 The systematic way to do these problems is as follows:
Step 1: Find the prime factorization, 1 8 1 8 = 2 × 3 2 × 1 0 1
Step 2: Write each prime p with an odd power as a sum of two squares, if you can; this is possible if and only if p = 2 or p ≡ 1 ( m o d 4 ) . In our case, 2 = 1 2 + 1 2 and 1 0 1 = 1 0 2 + 1 2
Step 3: Use the product formula, 2 0 2 = 1 0 1 × 2 = ( 1 0 2 + 1 2 ) ( 1 2 + 1 2 ) = ( 1 0 − 1 ) 2 + ( 1 0 + 1 ) 2 = 9 2 + 1 1 2 (this is also the formula Curtis used)
Step 4: Incorporate the even powers: 1 8 1 8 = 9 × 2 0 2 = 3 2 ( 1 1 2 + 9 2 ) = ( 3 × 1 1 ) 2 + ( 3 × 9 ) 2 = 3 3 2 + 2 7 2 .
It's always good to look for shortcuts, of course, as Curtis and Josh did, but it is also good to have a systematic approach to fall back on.
Log in to reply
it also comes from basic algebraic identity :
(a^{2} + b^{2} ) (c^{2} + d^{2} )
= a^{2} c^{2} + a^{2} d^{2}
+ b^{2} c^{2} + b^{2} d^{2}
= a^{2} c^{2} + b^{2} d^{2} +2abcd
+ b^{2} c^{2} + a^{2} d^{2} -2abcd
=(ac+bd)^{2} + (bc-ad)^{2}
To find the solution I noted that: 1 8 1 8 ≈ 2 × 3 0 2 Now from this I supposed what would happen if a was 'y' smaller than 30 and b was 'x' bigger than 30... ( 3 0 + x ) 2 + ( 3 0 − y ) 2 = 1 8 0 0 + 6 0 ( x − y ) + ( x 2 + y 2 ) = 1 8 1 8 ⟹ 1 8 = 6 0 ( x − y ) + ( x 2 + y 2 ) Now it is clear that x = y will produce a nice solution.. 2 x 2 = 1 8 ⟹ x = 3 ⇒ 2 7 2 + 3 3 2 = 1 8 1 8 ∴ a + b = 6 0
Nice creative solution! Great birthday present for Comrade Karl ;)
The short version of what you are saying is this: 1 8 1 8 = 2 × 3 0 2 + 2 × 3 2 = ( 3 0 + 3 ) 2 + ( 3 0 − 3 ) 2
Log in to reply
Is there an elegant solution if the line "has one and only one integer solution" is not mentioned (without listing out all the perfect squares less than 1818)?
Log in to reply
If there is an elegant solution in this special case, I'm sure you can find it! There is a theorem, of course: A solution is essentially unique (up to sign and order) iff n has at most one prime factor congruent to 1 mod 4.
I just wanted to honour my hero, Karl Marx, at the occasion of his birthday... I am delighted to see that this silly problem has generated some mathematical interest.
Nice creative solution! Great birthday present for Comrade Karl ;)
The short version of what you are saying is this: 1 8 1 8 = 2 × 3 0 2 + 2 × 3 2 = ( 3 0 + 3 ) 2 + ( 3 0 − 3 ) 2
@Otto Bretscher , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.
Log in to reply
It's Curtis' solution, condensed! I had a longer solution in mind, much "mathier", based on the prime factorization of 1818 (the "standard" way to do these kinds of problems).
Log in to reply
Oh great! Can you change your solution to your intended solution? Thank you.
Firstly. Notice that 1 8 1 8 = 1 0 0 × 1 8 + 1 × 1 8 = 1 0 1 × 1 8 . Using the Brahmagupta–Fibonacci identity, 1 8 1 8 = 1 0 1 × 1 8 = ( 1 0 2 + 1 2 ) ( 3 2 + 3 2 ) = ( 1 0 × 3 + 1 × 3 ) 2 + ( 1 0 × 3 − 1 × 3 ) 2 = 3 3 2 + 2 7 2
A quick bit of arithmetic gives 3 3 + 2 7 = 6 0